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2 assume the power series x n 0 a n z n has radius of

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2. Assume the power series X n =0 a n z n has radius of convergence equal to 2. Determine the radius of convergence of the series X n =0 a n z 2 n +1 . Solution. The new series can be written in the form z X n =0 a n ( z 2 ) n . It clearly converges at z if and only if the original series converges at z 2 . In particular, it converges for | z 2 | < 2; i.e., for | z | < 2, diverges for | z 2 | > 2; i.e., for | z | > 2. The radius is 2. 3. Let { a n } be a sequence of positive numbers such that n =1 a n < . Prove: There exists a sequence { c n } of positive numbers diverging to such that n =1 a n c n < . Solution. This is too nice an exercise to be spoiled already by a solution. Only two people selected it in the final and I am afraid that their “solutions” are totally wrong. Very wrong. Fundamentally wrong. Everybody still has a chance to work on it.
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4. Let F be an equicontinuous set of real valued functions on the interval [ a,b ]. That is, each element of F is a function from [ a,b ] to R and for every ² > 0 there is δ such that if x,y [ a,b ] and | x - y | < δ , then | f ( x ) - f ( y ) | < ² . Assume -∞ < a < b < . Prove: If the sequence of real numbers { f ( a ) : f ∈ F} is bounded, then the set F is uniformly bounded. Solution. I did this in class! (Maybe slurring over some details.) The hypothesis is that F is equicontinuous and { f ( a ) : f ∈ F} is bounded; let M > 0 be such that | f ( a ) | ≤ M for all f ∈ F . Let δ be the δ that works for ² = 1 in the definition of equicontinuity; that is, let δ > 0 be such that if x,y [ a,b ], then | f ( x ) - f ( y ) | < 1. Let n N be such that ( b - a ) /n < δ and let x 0 = a,x 1 = a + b - a n ,...,x k = a + k b - a n ,...,x n = x k = a + n b - a n = b. Notice x k - x k - 1 = ( b - a ) /n < δ for k = 1 , 2 ,...,n . Let x [ a,b ], f ∈ F . There is k ∈ { 1 ,...,n } such that x [ x k - 1 ,x k ], then | x - x k - 1 | < δ and we have: | f ( x ) | ≤ | f ( x ) - f ( a ) | + | f ( a ) | = | f ( a ) | + f ( x ) - f ( x k - 1 ) + k - 1 X j =1 [ f ( x j ) - f ( x j - 1 )] . If k = 1, we interpret 0 j =1 as 0; the sum does not appear. The sum, of course, telescopes to f ( x k - 1 ) - f ( x 0 ) = f ( x k - 1 ) - f ( a ). Thus | f ( x ) | ≤ | f ( a ) | + | f ( x ) - f ( x k - 1 ) | + k - 1 X j =1 | f ( x j ) - f ( x j - 1 ) | ≤ M + 1 + k - 1 X j =1 1 = M + k M + n. Since x [ a,b ] is arbitrary, we proved | f ( x ) | ≤ M + n for all x [ a,b ], f ∈ F . The sequence is uniformly bounded my M + n . 5. If f,g : R R define f g if and only if f = g a.e. Prove: is an equivalence relation. Solution.
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  • Spring '11
  • Speinklo
  • Empty set, Open set, Topological space, Dominated convergence theorem, Lebesgue integration, Non-measurable set

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2 Assume the power series X n 0 a n z n has radius of...

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